在表达式中让..使用警卫 [英] Using guards in let .. in expressions

问题描述

有时候我写这样的代码

solveLogic :: Int -> Int -> Int
solveLogic a b =
    let 
        x = 1
        brainiac
            | a >= x     = 1
            | a == b     = 333
            | otherwise  = 5
    in
        brainiac

每次我敦促编写不需要不必要的"brainiac"功能的东西时，就像这样:

And every time I have urge to write this things without unneeded "brainiac" function, like this:

solveLogic :: Int -> Int -> Int
solveLogic a b =
    let 
        x = 1
    in
        | a >= x     = 1
        | a == b     = 333
        | otherwise  = 5

哪个代码更像是"Haskellish".有什么办法吗?

Which code is much more "Haskellish". Is there any way of doing this?

推荐答案

是的，使用where子句:

solveLogic a b
        | a >= x     = 1
        | a == b     = 333
        | otherwise  = 5
    where
      x = 1

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